Moon.cfg in orbiter beta

[tblaxland]

I am not familiar with concept of Laplace plane but could we just compute mean value of axis of rotation over a precession cycle, shouldn't that be at least pretty close to nodal precession pole.

Yes it should, except that the precession cycle for the Moon is only 18.6 years and there are longer period variations. I'll start by considering J2000 +/- 100 years and see what it looks like. I haven't had a chance to look at this since last week but hopefully sometime in the next week I will be able to get to it.

Regarding the Laplace plane: There are two types of precession at work when we consider the "precession of the Moon". One is the precession of the equatorial plane and the other is the precession of the orbital plane. The Laplace plane is normal to the axis about which the orbital plane precesses. In the case of the Moon, its Laplace plane is coincident with the ecliptic.

 

[tblaxland]

jarmonik, I should add to my last post that Orbiter does not currently have the capability to model the precession of the orbital plane via the config file (Martin, I expect that you have no intention of providing this capability at this stage, if at all?). The precession of the Moon's orbital plane is presumably modelled in its module (I guess the data in the elp82.dat file contains its position). I think you would have noticed if this was not modelled adequately in your work with the AGC.

Edit: Also, I am beginning to realise that L_{ecl} (longitude with respect to ecliptic frame) in equation 7 is a mistake. The rotation should be independent of the inertial frame of the observer, so one can always select a frame in which \eps_{ref} = 0. Therefore, L_{ecl} should be replaced by L_{rel} in Eq. 7.

I have finally had a chance to review this document and agree with this change.

Now to see if I can confirm jarmonik's values for the Moon from the IAU document...

 

[jarmonik]

I think you would have noticed if this was not modelled adequately in your work with the AGC.

I haven't done any work with the AGC and I have no part in that project.

Regarding the Laplace plane: There are two types of precession at work when we consider the "precession of the Moon". One is the precession of the equatorial plane and the other is the precession of the orbital plane. The Laplace plane is normal to the axis about which the orbital plane precesses. In the case of the Moon, its Laplace plane is coincident with the ecliptic.

That's true, but if the Laplace plane is a valid reference plane for precession of the orbit, then why wouldn't it be a valid reference for precession of the equator. It would serve the same purpose in both cases. Of course, the orientation of these planes wouldn't be the same unless the system is in Cassini state.

 

[tblaxland]

I haven't done any work with the AGC and I have no part in that project.

My apologies. When you mentioned differences between the AGC and IAU models earlier, I misunderstood your involvement.

That's true, but if the Laplace plane is a valid reference plane for precession of the orbit, then why wouldn't it be a valid reference for precession of the equator. It would serve the same purpose in both cases.

Any inertial frame would be a valid reference, so yes, the Laplace plane is a valid reference. I don't think I have said it is not. But since the Laplace plane is never defined in a body's cfg file, it makes sense to use the ecliptic as the reference instead.

Of course, the orientation of these planes wouldn't be the same unless the system is in Cassini state.

By "these planes" you mean the Laplace plane and a plane normal to the axis of precession of the equatorial plane (I don't know of a special name for that plane...)? I agree.

 

[Chode]

I've noticed an inconsistency between the description of how rotations are calculated and the actual mathematics that Orbiter uses. You can see it in Figure 1 of the technote, which would imply that for LAN = 0, a positive obliquity would result in the pole vector having a negative z component, where the rotation matrices actually give a positive z component (multiplying (0,1,0) by the rotation matrix). This is due to the fact that all the descriptions of rotation calculations in Orbiter state that LAN is the angle from the vernal equinox to the ascending node of the equator with respect to the ecliptic, while the math actually takes LAN as meaning the angle to the descending node. This is most easily seen by considering the vernal equinox, which in the real world is defined as the ascending node of the ecliptic with respect to the Earth's equator, which is actually the descending node of the Earth's equator w.r.t. the ecliptic.

For practical purposes, all this means is that the LAN that you need to give Orbiter to get the physically correct rotations is 180 degrees off from the LAN of the equator w.r.t. the ecliptic.

Also, I think that the transform matrix is more accurately called the matrix that transforms from local to ecliptic coordinates.

Regards

 

[tblaxland]

You can see it in Figure 1 of the technote, which would imply that for LAN = 0, a positive obliquity would result in the pole vector having a negative z component, where the rotation matrices actually give a positive z component (multiplying (0,1,0) by the rotation matrix).

The rotation matrix for inclination is, in fact, correct. You must bear in mind that the matrix is for a reference frame transformation not a vector rotation, so it is the transpose of the normal rotation matrix. So consider a case where we have LAN=0 and the vector (0,1,0) in the ecliptic frame. Multiplying by the rotation matrix yields positive y & z values for an inclination of between 0 and +90. Referring back to Figure 1, the ecliptic north pole does indeed have positive y & z values in the precession axis reference frame.

Having said that, I think there is an error in the matrix for the LAN transformation. Consider a case of LAN=90, e=0 and a vector (1,0,0) in the ecliptic frame. Multiplying by the matrix given in equation 1, you get (0,0,1) in the precession axis frame, yet this should be (0,0,-1). So I think that in equations 1 & 3, the first matrix should be transposed.

 

[Chode]

Having said that, I think there is an error in the matrix for the LAN transformation. Consider a case of LAN=90, e=0 and a vector (1,0,0) in the ecliptic frame. Multiplying by the matrix given in equation 1, you get (0,0,1) in the precession axis frame, yet this should be (0,0,-1). So I think that in equations 1 & 3, the first matrix should be transposed.

It depends on what this transformation is supposed to be. I think it should be the transformation from local to ecliptic coordinates, since in the end, you multiply it by the pole vector in local coords (0,1,0) to get the obliquity and LAN with respect to the ecliptic. If that is the case, then to agree with the diagram, you would need to transpose the obliquity matrix. If, instead, you want the matrix to represent a transformation from ecliptic to local coordinates, then you would transpose the L matrix, and change the order. Those two transformation matrices would then be the inverse (=transpose) of each other, which is what they should be.

What I am suggesting is that neither should be done, but instead recognize that what Orbiter calls LAN is actually the angle to the descending node of the equator with respect to the ecliptic, since the vernal equinox is the descending node of the Earth's equator w.r.t. the ecliptic. Then the only thing that would change is the diagram.

Regards

 

[tblaxland]

OK, I agree with on this: "the vernal equinox is the descending node of the Earth's equator w.r.t. the ecliptic" but disagree on this: "what Orbiter calls LAN is actually the angle to the descending node of the equator with respect to the ecliptic" because I think the LAN of Earth's equatorial frame should be 180 deg.

I also agree on this: "I think it should be the transformation from local to ecliptic coordinates, since in the end, you multiply it by the pole vector in local coords (0,1,0) to get the obliquity and LAN with respect to the ecliptic". This is required in order to make equation 6 correct.

 

[Chode]

but disagree on this: "what Orbiter calls LAN is actually the angle to the descending node of the equator with respect to the ecliptic" because I think the LAN of Earth's equatorial frame should be 180 deg.

I agree that the LAN of Earth's equator w.r.t. the ecliptic is 180 degrees in reality, but if you want to have that value in the config file, then Orbiter math would need to be changed, and then the config files for all celestial bodies would also need to reflect that change.

Regards

---------- Post added at 07:45 PM ---------- Previous post was at 04:06 PM ----------

In my opinion, the simplest fix for all this is to change the documentation, so that LAN is described as "the angle from the vernal equinox to the ascending node of the ecliptic w.r.t. the body equator". This would also mean that the figs in the technote be changed accordingly.

Regards

 

[tblaxland]

In my opinion, the simplest fix for all this is to change the documentation, so that LAN is described as "the angle from the vernal equinox to the ascending node of the ecliptic w.r.t. the body equator". This would also mean that the figs in the technote be changed accordingly.

Yes, makes sense.

 

[martins]

Thanks for the review, guys.

I agree that this needs to be fixed. I'll update the documentation and technote for the next beta.

 

[jarmonik]

There is something wrong. Geocentric longitude shown by SurfaceMFD and LunarTransferMFD is not the same when PrecessionObliquity and
PrecessionLAN values are given for the Moon. (This is just a test).

I don't know witch one of these values is correct.

Since, the Latitude is the same in both MFDs it would give a reason to suspect that something might be wrong in oapiGetPlanetCurrentRotation()

When PrecessionObliquity and PrecessionLAN are removed, both MFDs will show the same results.

I'll do some more testing.

---------- Post added at 10:09 PM ---------- Previous post was at 08:53 PM ----------

In my current test scenario value returned by oapiGetPlanetCurrentRotation() is about -18.7

 

[tblaxland]

Now to see if I can confirm jarmonik's values for the Moon from the IAU document...

Getting back to this, I was looking for info on converting from ICRF to J2000 ecliptic/equinox and I stumbled across Allen's astrophysical quantities, courtesy of Google Books:
http://books.google.com/books?id=w8...ontcover&source=gbs_summary_r&cad=0#PPA294,M1

I found the info I needed on pg 294.

 

[jarmonik]

Getting back to this, I was looking for info on converting from ICRF to J2000 ecliptic/equinox and I stumbled across Allen's astrophysical quantities, courtesy of Google Books:
http://books.google.com/books?id=w8...ontcover&source=gbs_summary_r&cad=0#PPA294,M1

I found the info I needed on pg 294.

I am not sure is that correct. Also the reference plane of lunar model may not be the ICRF.

This document http://ssd.jpl.nasa.gov/dat/lunar_cmd_2005_jpl_d32296.pdf contains the same numbers for lunar model and they do not speak of ICRF. Except in a side note in page 36

Scientific documentation shouldn't leave that much room for speculation.

 

[tblaxland]

Not sure the values are correct?

The IAU/IAG Working Group report specifies that it uses the ICRF. It also gives what it calls "approximate values" for the invariable plane in the ICRF - these correlate with Allen's to within the precision given given in their report.

Do you have alternative data that relates the ICRF to the J2000 ecliptic/equinox? I couldn't find anything on the IERS website (the people that maintain the ICRF).

 

[jarmonik]

Do you have alternative data that relates the ICRF to the J2000 ecliptic/equinox?

That depends from the question below.

What's the difference between "J2000 ecliptic/equinox" and "Earth Mean Equator and Equinox of J2000" ?

The difference between ICRF and "Earth Mean Equator and Equinox of J2000" is a few milli-arcseconds.

 

[tblaxland]

What's the difference between "J2000 ecliptic/equinox" and "Earth Mean Equator and Equinox of J2000" ?

The difference would be the mean obliquity of the ecliptic with respect to the Earth's equator.

The difference between ICRF and "Earth Mean Equator and Equinox of J2000" is a few milli-arcseconds.

In that case, I just need good data on the mean obliquity of the ecliptic. I can calculate that from Allen's. If you would be so kind as to provide a reference I could compare the two. I wouldn't have thought this such a hard thing to find but Googling around I find nothing I would consider "authoritative".

 

[jarmonik]

There are some information in page 36 in a docucment I linked few post above. The information comes from this document http://articles.adsabs.harvard.edu/...IEW&data_type=PDF_HIGH&send=GET&filetype=.pdf

There are several sources for Mean Obliquity but non very reliable.

It looks like the Mean Obliquity is 23° 26' 21.448'' http://www.geoastro.de/obliquity/index.html the same information can be found from wikipedia.

I'll search more sources...

But where does the LAN 3°51' (in Allen's) come from ?

 

[tblaxland]

But where does the LAN 3°51' (in Allen's) come from ?

Are you sure you are not getting confused between the ecliptic and the invariable plane? 3°51' is the LAN of invariable (aka Laplacian) plane, ie, the plane normal to the total relative orbital angular momentum of the solar system (excluding the sun). EDIT: I'm not sure there is an answer to "where does it come from?" - I think that falls into the philosophical realm... I'll use the answer I sometimes have to give my 4yo son - "It just Is"

I was going to convert from ICRF to the invariable plane, then from the invariable plane to the ecliptic but it is obviously easier to go direct if the obliquity of the ecliptic in the ICRF is known (or even in Mean Equator and Equinox of J2000 - same as FK5, yes?). I'll work on that assumption for now and use the value you quoted. I can update it easily enough if a better value comes along.

---------- Post added at 18:18 ---------- Previous post was Yesterday at 21:39 ----------

OK, I started by looking at Earth's parameters from the IAU/IAG document and I am confused because my SideRotOffset looks like it is out by -90 degrees from what is in the beta config file (I have 3.3186912). For the ICRF, where is the origin of the right ascensions? The way it is described in the IAU/IAG document is that the intersection of the ICRF and the ecliptic should occur at right ascension 90 and 270.

 

[jarmonik]

Are you sure you are not getting confused between the ecliptic and the invariable plane?

Yes, I did confuse it to ecliptic. Wouldn't be bad idea to use that data to double check the orientation of ICRF. However, the difference between ICRF and EME2000 (Earth Mean Equator) is so small that it probably doesn't matter in the Orbiter.

(or even in Mean Equator and Equinox of J2000 - same as FK5, yes?).

I don't know.

OK, I started by looking at Earth's parameters from the IAU/IAG document and I am confused because my SideRotOffset looks like it is out by -90 degrees from what is in the beta config file (I have 3.3186912). For the ICRF, where is the origin of the right ascensions?

Based on the documentation, the origin of right ascension should be the vernal equinox. (Intersection of ICRF and ecliptic)

The way it is described in the IAU/IAG document is that the intersection of the ICRF and the ecliptic should occur at right ascension 90 and 270

No, that would be 0 and 180

PS. Let's remember that the difference between the Earth's equator and ICRF is very small. It seems that the LAN of the Earth's equator respect to ICRF is 90 degrees and the inclination is zero (in the epoch of J2000). Where as the LAN of the Earth's equator respect to ecliptic is 0 degrees and the inclination is 23.5

---------- Post added at 05:53 PM ---------- Previous post was at 01:51 PM ----------

OK, I started by looking at Earth's parameters from the IAU/IAG document and I am confused because my SideRotOffset looks like it is out by -90 degrees from what is in the beta config file (I have 3.3186912).

I haven't computed any values but that sounds like it's correct. Note that LAN has changed 90 degrees as well.